1. Center of Mass & Linear Momentum
Unit 4

2. The center of mass (com) of an object is the point at which all of the object’s mass could be considered to be located
The com of a symmetrical uniform object is located at its geometrical center
The com of an object does NOT have to lie within the object
A system of objects also has a com

3. The x-coordinate for com is given by:
𝑥 𝑐𝑜𝑚 = 𝑚 1 𝑥 1 + 𝑚 2 𝑥 2 +… 𝑚 𝑛 𝑥 𝑛 𝑚 1 + 𝑚 2 +… 𝑚 𝑛
The y-coordinate would work the same way
𝑦 𝑐𝑜𝑚 = 𝑚 1 𝑦 1 + 𝑚 2 𝑦 2 +… 𝑚 𝑛 𝑦 𝑛 𝑚 1 + 𝑚 2 +… 𝑚 𝑛

4. Determine the x and y coordinates for com
𝑥 𝑐𝑜𝑚 =6.27
𝑦 𝑐𝑜𝑚 =7.09

5. Determine the x and y coordinates for com
𝑥 𝑐𝑜𝑚 =3.84
𝑦 𝑐𝑜𝑚 = 3.16

6. Newton’s 2nd Law applies to com
𝐹 𝑛𝑒𝑡 = 𝑀𝑎 𝑐𝑜𝑚

7. Momentum is the vector product of mass and velocity
𝑝=𝑚𝑣
𝐹= 𝑑𝑝 𝑑𝑡

8. The change in an object’s momentum is known as impulse (J)
𝐽=Δ𝑝= 𝑝 𝑓 − 𝑝 𝑖 =𝑚 𝑣 𝑓 − 𝑣 𝑖
𝐽= 𝑡 𝑖 𝑡 𝑓 𝐹 𝑡 𝑑𝑡
𝐽= 𝐹 𝑎𝑣𝑔 Δ𝑡

9. In the absence of a net external force, momentum is conserved
𝑝 𝑓 = 𝑝 𝑖
Even with a net external force, momentum can be conserved.
In directions perpendicular to the force

10. Types of collisions
Elastic
Inelastic
Perfectly inelastic

11. Elastic collisions Objects bounce off No deformation
Momentum is conserved
Energy is conserved

12. Inelastic collisions Objects bounce off Possible deformation
Momentum is conserved
Energy is NOT conserved

13. Perfectly inelastic collisions
Objects stick together
Possible deformation
Momentum is conserved
Energy is NOT conserved

14. A 0.75 kg ball is moving east at 25 m/s when it strikes a vertical wall and rebounds at 20 m/s to the west. Determine the ball’s change in momentum
∆𝑝= 𝑝 𝑓 − 𝑝 𝑖
∆𝑝= 𝑚𝑣 𝑓 − 𝑚𝑣 𝑖 =𝑚 𝑣 𝑓 − 𝑣 𝑖
∆𝑝=0.75 −20−25
∆𝑝=−33.8 kg∙m/s or N∙s

15. When projectile m1 travelling at v1 hits stationary target m2 in an elastic collision
𝑚 1 𝑣 1𝑖 = 𝑚 1 𝑣 1𝑓 + 𝑚 2 𝑣 2𝑓
1 2 𝑚 1 𝑣 1𝑖 2 = 1 2 𝑚 1 𝑣 1𝑓 𝑚 2 𝑣 2𝑓 2
𝑚 1 𝑣 1𝑖 2 = 𝑚 1 𝑣 1𝑓 2 + 𝑚 2 𝑣 2𝑓 2
𝑚 1 𝑣 1𝑖 − 𝑚 2 𝑣 2𝑓 = 𝑚 1 𝑣 1𝑓
𝑚 1 𝑣 1𝑖 − 𝑚 2 𝑣 2𝑓 𝑚 1 2 = 𝑣 1𝑓 2

16. 𝑚 1 𝑣 1𝑖 − 𝑚 2 𝑣 2𝑓 𝑚 1 2 = 𝑣 1𝑓 2
𝑚 1 2 𝑣 1𝑖 2 −2 𝑚 1 𝑚 2 𝑣 1𝑖 𝑣 1𝑓 + 𝑚 2 2 𝑣 2𝑓 2 𝑚 1 2 = 𝑣 1𝑓 2
𝑚 1 𝑣 1𝑖 2 = 𝑚 1 𝑣 1𝑓 2 + 𝑚 2 𝑣 2𝑓 2
𝑚 1 𝑣 1𝑖 2 = 𝑚 1 𝑚 1 2 𝑣 1𝑖 2 −2 𝑚 1 𝑚 2 𝑣 1𝑖 𝑣 2𝑓 + 𝑚 2 2 𝑣 2𝑓 2 𝑚 𝑚 2 𝑣 2𝑓 2
𝑚 1 2 𝑣 1𝑖 2 = 𝑚 1 2 𝑣 1𝑖 2 −2 𝑚 1 𝑚 2 𝑣 1𝑖 𝑣 2𝑓 + 𝑚 2 2 𝑣 2𝑓 2 + 𝑚 1 𝑚 2 𝑣 2𝑓 2
0=−2 𝑚 1 𝑚 2 𝑣 1𝑖 𝑣 2𝑓 + 𝑚 2 2 𝑣 2𝑓 2 + 𝑚 1 𝑚 2 𝑣 2𝑓 2
2 𝑚 1 𝑚 2 𝑣 1𝑖 𝑣 2𝑓 = 𝑚 2 2 𝑣 2𝑓 2 + 𝑚 1 𝑚 2 𝑣 2𝑓 2

17. 2 𝑚 1 𝑚 2 𝑣 1𝑖 𝑣 2𝑓 = 𝑚 2 2 𝑣 2𝑓 2 + 𝑚 1 𝑚 2 𝑣 2𝑓 2
2 𝑚 1 𝑚 2 𝑣 1𝑖 = 𝑚 2 2 𝑣 2𝑓 + 𝑚 1 𝑚 2 𝑣 2𝑓
2 𝑚 1 𝑚 2 𝑣 1𝑖 = 𝑚 𝑚 1 𝑚 2 𝑣 2𝑓
2 𝑚 1 𝑚 2 𝑣 1𝑖 𝑚 𝑚 1 𝑚 2 = 𝑣 2𝑓
𝑣 2𝑓 = 2 𝑚 1 𝑣 1𝑖 𝑚 2 + 𝑚 1

18. 𝑣 2𝑓 = 2 𝑚 1 𝑣 1𝑖 𝑚 2 + 𝑚 1
𝑚 1 𝑣 1𝑖 = 𝑚 1 𝑣 1𝑓 + 𝑚 2 𝑣 2𝑓
𝑚 1 𝑣 1𝑖 = 𝑚 1 𝑣 1𝑓 + 𝑚 2 2 𝑚 1 𝑣 1𝑖 𝑚 2 + 𝑚 1
𝑣 1𝑖 = 𝑣 1𝑓 + 𝑚 2 2 𝑣 1𝑖 𝑚 2 + 𝑚 1
𝑣 1𝑓 = 𝑣 1𝑖 − 2 𝑚 2 𝑣 1𝑖 𝑚 2 + 𝑚 1

19. What if…
m1 = m2?
𝑣 1𝑓 = 𝑣 1𝑖 − 2 𝑚 2 𝑣 1𝑖 𝑚 2 + 𝑚 1
𝑣 2𝑓 = 2 𝑚 1 𝑣 1𝑖 𝑚 2 + 𝑚 1
𝑣 1𝑓 = 𝑣 1𝑖 − 2 𝑚 2 𝑣 1𝑖 𝑚 2 + 𝑚 2 = 𝑣 1𝑖 − 2 𝑚 2 𝑣 1𝑖 2 𝑚 2 = 𝑣 1𝑖 − 𝑣 1𝑖 =0
𝑣 2𝑓 = 2 𝑚 1 𝑣 1𝑖 𝑚 2 + 𝑚 1 = 2 𝑚 1 𝑣 1𝑖 𝑚 1 + 𝑚 1 = 2 𝑚 1 𝑣 1𝑖 2𝑚 1 = 𝑣 1𝑖

20. What if…
m1 << m2?
𝑣 1𝑓 = 𝑣 1𝑖 − 2 𝑚 2 𝑣 1𝑖 𝑚 2 + 𝑚 1
𝑣 2𝑓 = 2 𝑚 1 𝑣 1𝑖 𝑚 2 + 𝑚 1
𝑣 1𝑓 ≈ 𝑣 1𝑖 − 2 𝑚 2 𝑣 1𝑖 𝑚 2 ≈ 𝑣 1𝑖 −2 𝑣 1𝑖 ≈− 𝑣 1𝑖
𝑣 2𝑓 = 2 𝑚 1 𝑣 1𝑖 𝑚 2 + 𝑚 1 ≈ 2 𝑚 1 𝑣 1𝑖 𝑚 2 ≈0

21. What if…
m1 >> m2?
𝑣 1𝑓 = 𝑣 1𝑖 − 2 𝑚 2 𝑣 1𝑖 𝑚 2 + 𝑚 1
𝑣 2𝑓 = 2 𝑚 1 𝑣 1𝑖 𝑚 2 + 𝑚 1
𝑣 1𝑓 ≈ 𝑣 1𝑖 − 2 𝑚 2 𝑣 1𝑖 𝑚 1 ≈ 𝑣 1𝑖 −0≈ 𝑣 1𝑖
𝑣 2𝑓 ≈ 2 𝑚 1 𝑣 1𝑖 𝑚 1 ≈2 𝑣 1𝑖